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It is given that in a triangle ABC, a line from A to BC intersects BC at point D. If the ratio in which AD divides BC is given can we say anything about the ratio of areas of triangle ABD and triangle ADC ?

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Yes we can find the ratio of the areas of the two triangles.
Area of $\triangle ADC$=$$(AD\times DC\times\sin\theta)\over2$$ Area of $\triangle ABD$=$$(AD\times DB\times \sin(180^\circ-\theta))\over2$$ [$\theta$ is the angle $ADC$]

Divide these two equations and use the ratio you have to get the answer.

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    $\begingroup$ So basically the areas are divided in the ratio of the lengths of AD and DB as the rest of the terms will cancel. $\endgroup$ – Suy Oct 10 '13 at 18:38
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    $\begingroup$ Yes that is true. $\endgroup$ – iajnr Oct 10 '13 at 18:39
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    $\begingroup$ More simply, this results from the fact that both triangles have the same altitude $\endgroup$ – Bill Kleinhans Oct 10 '13 at 20:42

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